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#### The Fundamental Theorem of Calculus

Three Different Quantities
The Whole as Sum of Partial Changes
The Indefinite Integral as Antiderivative
The FTC and the Chain Rule

#### The Indefinite Integral and the Net Change

Indefinite Integrals and Anti-derivatives
A Table of Common Anti-derivatives
The Net Change Theorem
The NCT and Public Policy

#### Substitution

Substitution for Indefinite Integrals
Revised Table of Integrals
Substitution for Definite Integrals

#### Area Between Curves

The Slice and Dice Principle
To Compute a Bulk Quantity
The Area Between Two Curves
Horizontal Slicing
Summary

#### Volumes

Slicing and Dicing Solids
Solids of Revolution 1: Disks
Solids of Revolution 2: Washers
Volumes Rotating About the $y$-axis

Behind IBP
Examples
Going in Circles

#### Integrals of Trig Functions

Basic Trig Functions
Product of Sines and Cosines (1)
Product of Sines and Cosines (2)
Product of Secants and Tangents
Other Cases

#### Trig Substitutions

How it works
Examples
Completing the Square

#### Partial Fractions

Introduction
Linear Factors
Improper Rational Functions and Long Division
Summary

#### Strategies of Integration

Substitution
Integration by Parts
Trig Integrals
Trig Substitutions
Partial Fractions

#### Improper Integrals

Type I Integrals
Type II Integrals
Comparison Tests for Convergence

#### Differential Equations

Introduction
Separable Equations
Mixing and Dilution

#### Models of Growth

Exponential Growth and Decay
Logistic Growth

#### Infinite Sequences

Close is Good Enough (revisited)
Examples
Limit Laws for Sequences
Monotonic Convergence

#### Infinite Series

Introduction
Geometric Series
Limit Laws for Series
Telescoping Sums and the FTC

#### Integral Test

The Integral Test
When the Integral Diverges
When the Integral Converges

#### Comparison Tests

The Basic Comparison Test
The Limit Comparison Test

#### Convergence of Series with Negative Terms

Introduction
Alternating Series and the AS Test
Absolute Convergence
Rearrangements

The Ratio Test
The Root Test
Examples

#### Strategies for testing Series

List of Major Convergence Tests
Examples

#### Power Series

Finding the Interval of Convergence
Other Power Series

#### Representing Functions as Power Series

Functions as Power Series
Derivatives and Integrals of Power Series
Applications and Examples

#### Taylor and Maclaurin Series

The Formula for Taylor Series
Taylor Series for Common Functions
Adding, Multiplying, and Dividing Power Series
Miscellaneous Useful Facts

#### Applications of Taylor Polynomials

What are Taylor Polynomials?
How Accurate are Taylor Polynomials?
What can go Wrong?
Other Uses of Taylor Polynomials

#### Partial Derivatives

Definitions and Rules
The Geometry of Partial Derivatives
Higher Order Derivatives
Differentials and Taylor Expansions

#### Multiple Integrals

Background
What is a Double Integral?
Volumes as Double Integrals

#### Iterated Integrals over Rectangles

One Variable at the Time
Fubini's Theorem
Notation and Order

#### Double Integrals over General Regions

Type I and Type II regions
Examples
Order of Integration
Area and Volume Revisited

### Examples

 Examples 1, 2: In the following video we integrate $f(x,y) = x^2y$ over the upper half of the unit disk; and $f(x,y) = 4xe^{2y}$ over the region bounded by the $x$-axis, the $y$-axis, the line $y=2$ and the curve $y=\ln(x)$.

 Example 3: Evaluate the integral $$I \ = \ \int \int_D\, (x+y)\, dA$$ when $D$ consists of all points $(x,\,y)$ such that $$0 \ \le \ y \ \le \sqrt{9-x^2}\,, \quad 0\ \le \ x \ \le 3\,.$$ Solution: Since $y^2 \ = \ 9 - x^2$ is a circle of radius $3$ centered at the origin, $D$ consists of all points in the first quadrant inside this circle as shown below. This is a Type I region, which suggests Fixing $x$ and integrating with respect to $y$ along the black vertical line as shown, and Then integrating with respect to $x$. So the double integral $I$ becomes the repeated integral $$\int_0^3\left(\int_0^{\sqrt{9-x^2}}\, (x+y)\, dy\right)\,dx= \int_0^3\, \Bigl[\, xy - \frac{1}{2}y^2 \,\Bigl]_0^{\sqrt{9 - x^2}}\,dx.$$ Thus $$I\ = \ \int_0^3\, \Bigl( x\sqrt{9 - x^2} - \frac{1}{2} (9 - x^2)\Bigr)\, dx\ = \ 18\,.$$ How would you evaluate this last integral?

In Example 3 algebraic conditions specifying $D$ suggested how to write the integral as a repeated integral. Other times algebraic conditions are best interpreted graphically before deciding on limits of integration.

 Example 4: Evaluate the integral $$I \ = \ \int \int_D\, (3x +4y)\, dA$$ when $D$ is the bounded region enclosed by $y = x$ and $y=x^2$. Here $D$ is enclosed by the straight line $y = x$ and the parabola $y = x^2$ as shown below. To determine the limits of integration we first need to find the points of intersection of $y = x$ and $y = x^2$. These occur when $x^2 = x$, i.e., when $x = 0,\, 1$. Treating $D$ as a Type I region, we fix $x$ and integrate with respect to $y$ along the black vertical line, getting the repeated integral $$I \ = \ \int_0^1\left(\int_{x^2}^{x}\, (3x + 4y)\, dy\right)\,dx$$ Can you now evaluate $I$?

In Example 4 it was simpler to fix $x$ and integrate first with respect to $y$ because the bounding curves were already given to us as $$y \ = \ f_1(x)\ = \ x^2\,,\qquad y \ = \ f_2(x)\ = \ x\,.$$ Had they been given as $x \,=\, g_1(y)$ and $x\,=\, g_2(y)\,$ it might have been easier to think of $D$ as a Type II region. We would fix $y$ and first integrate with respect to $x$.

 Example 5: The region $D$ that is shown below is Type I but is not Type II. Fixing $x$ and integrating first with respect to $y$ along the black line makes good sense because then $$D \ = \ \Bigl\{\,(x,\,y) : \phi(x) \le y \le \psi(x),\ \ a \le x \le b\,\Bigl\}$$ for suitable choices of $a,\, b$ and functions $\phi(x),\, \psi(x)$: $$\int \int_D\, f(x,\,y)\, dxdy = \int_a^b \left(\int_{\phi(x)}^{\psi(x)}\, f(x,\,y)\, dy\right) dx\,.$$ But if we had chosen to fix $y$, then the integral with respect to $x$ would sometimes split into two parts shown in red. Not a good idea!

 Example 6: The region $D$, shown below, is Type II but not Type I. Fixing $y$ and integrating first with respect to $x$ along the black line makes good sense because then $$D \ = \ \Bigl\{\,(x,\,y) : \phi(y) \le x \le \psi(y),\ \ c \le y \le d\,\Bigl\}$$ for suitable choices of $c,\, d$ and functions $\phi(y),\, \psi(y)$. In this case $$\int \int_D\, f(x,\,y)\, dxdy = \int_c^d \left(\int_{\phi(y)}^{\psi(y)}\, f(x,\,y)\, dx\right) dy\,.$$ But if we had chosen to fix $x$, then the integral with respect to $y$ would sometimes splits into two parts as shown in red. Again not a good idea!